Here's the question you clicked on:
aroub
Why do we have to put an absolute value in the final answer?!
\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}\]
Can't even see that, is it 10 th root?
see moderator has warned me now...whats the way to delet ur account...can somebody give the link...i m going frm here
can u pls give me the link...i want to delet my account..i can't find it anywhere
I don't know.. You have to contact a moderator
cmon moderator show up now...
The only reason I can think of is they want to make sure you get a real and postiive (as opposed to complex) root. There will be 10 different roots.
@akash_809 you cannot delete OS accounts
But you can write it without absolute value, right? Because (or this is what I've been taught) there's x^3 in the radicand so it's definitely positive..
the simple reason is that\[x^{10}=(-x)^{10}\]so \[\large\sqrt[10]{x^{10}}\]will give you the same answer whether x is negative or positive
what...thats cynical...open study community is surely going nowher with all these warnings and suspensions...and to top that off , you even cant delete your account...what is this open study team please response...i wnat to know how can i delete my account
absolute value of x is actually defined in 2 ways: 1) distance from the point x=0 on the number line 2) \(|x|=\sqrt{x^2}\)
so since\[\sqrt{(-x)^2}=\sqrt{x^2}=|x|\]then\[\sqrt[10]{x^{10}}=\sqrt[10]{(-x)^{10}}=|x|\](because x to any even power is positive, whether x>0 or x<0)
does that help you @aroub ?
In this case I totally understand why. But in the question itself there's an odd exponent in the radicand so it's definitely positive right? So then no need for an absolute value..
@TuringTest root of negative number is not defined...it will result in iota value...so -x wont be the answer i think
not necessarily, let's assume x<0...
If you assumed that, the question will be wrong.
yes it is...\[lets take an example......\sqrt{-2} is \not defined unless you are taking imaginary numbers into account then u wl get +-1.414 i\]
the root of a negative number is not undefined, it is imaginary...\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}\]
sry reply got messed up i was just saying that square root of -2 is not defined unless you consider imaginary numbers as wel..
Put imaginary numbers on a side for now..
imaginary numbers are defined... now if x<0, then x^3<0 that means that (x^3)(x^3)>0 which is what we would expect given that x^6>0 since all numbers to a positive exponent are >0
another way to see it, let's assume x>0\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}=\sqrt[10]{x^{10}}=|x|\]now we do it with the negative of x\[\sqrt[10]{(-x)^4} \times \sqrt[10]{(-x)^3} \times \sqrt[10]{(-x)^3}=\sqrt[10]{(-x)^4}\times\sqrt[10]{(-x)^3\cdot (-x)^3}\]\[=\sqrt[10]{(-x)^{10}}=\sqrt[10]{x^{10}}=|x|\]so you see, after all is said and done we do not know whether the x we had at the beginning was positive or negative, hence we can only conclude that the answer is |x|
Okay take this as another example: \[4\sqrt[4]{5a^2}\times 8\sqrt[4]{125a}\times 2\sqrt[4]{a}\] Here, a does not need an absolute value. Why? You can assume that a>0
yes you would have to assume a >0
let's just look at powers of a...
if a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a\](since we already knew a>0 now with the negative of a...
yea i m waiting for that negative values of a...what would it give
Aoh, so you have to try the negative and the positive?
\[\sqrt[4]{(-a)^2}\times\sqrt[4]{-a}\times\sqrt[4]{-a}=\sqrt[4]{a^4}=|a|\neq a\]because |a|>0, but we just assumed a<0 ! a contradiction, hence the assumption makes a difference
sorry, I mean we took the negative of a...
let me write that again, gotta be careful here
that's the problem...the thing you did here is called fallacy of maths... http://en.wikipedia.org/wiki/Mathematical_fallacy ....search invalid roots in this article
assume a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a\]assume a<0, then\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|\neq a\]since we assumed a<0
@akash_809 yes but at least one of these numbers is positive because it is raised to an even power, so I have committed no fallacy
did u went through the link......the proofs given like \[1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i.i=-1 ....so u hv 1=-1...and certainly \it is \not valid..\]
same thing with the earlier problem...
and 1=-1....surely not right...
yes, but\[\sqrt{(-1)^2}\cdot\sqrt{-1}=\sqrt1\cdot\sqrt{-1}\] is not the same as\[\sqrt{-1}\cdot\sqrt{-1}\]
notice we have a^2 in the above, so we know *that* part is positive
this turned out to be more subtle than I bet you realized @aroub :)
lol, yeah true xD ALWAYS thought absolute values are so easy! And whenever someone asks me about absolute values I answer until this question came up -.-' This question is a bit confusing >.< Well, THANKS LOOAADSSSSSS :D